Int Nature Ans 07

1

New Zealand Institute of Physics

ASSESSMENT SCHEDULE

Physics Level 2: 2007 90258 v2 Demonstrate understanding of physics in an integrated context

Note: Minor computational errors will not be penalised. A wrong answer will be accepted as correct provided there is sufficient evidence that the mistake is not due to a lack of understanding. Such evidence includes: • •

the last written step before the answer is given has no unexpanded brackets or terms and does not require rearranging. the power of any number that is multiplied by a power of 10 is correct.

Correct units and significant figures are required only in the questions that specifically ask for them.

90258 • Demonstrate understanding of physics in an integrated context • Assessment Schedule 2007

2 Evidence Statement Question

Evidence

Achievement

u&sf

Q1a(ii) rounded to 2 sf

One (a)

vvertical = 38 × tan 4.5 = 2.99066

(b)

(c)

(d)

Excellence

Correct Correct working

The only force acting on the ball is gravity and so it has constant horizontal speed and constant vertical acceleration. These are the required conditions for projectile motion. d = vit + ½at2 = (2.991× 0.47) + ½ × 10 × (0.47)2 =2.51027 = 2.5 m

Only force acting is gravity

The strings gain elastic potential energy as they stretch from the kinetic energy of the arm which came from the chemical potential energy in the arm muscles of the player. This elastic potential energy is changed to kinetic energy of the ball as the strings regain their normal length.

Idea of elastic potential energy

(e) (f)

Merit

Correct conditions, correct reason Correct answer

Idea of elastic potential energy being changed to kinetic energy

Explanation is clear and complete

Correct direction ∆ p = Ft ⇒t =

Correct answer

2. 3 410

−

= 5.609756× 10 3

−

= 5.6× 10 3 s (g)

∆ p = pf − pi − ⇒ 2.3 = 23× m − 17× m ⇒ 2.3 = 40× m ⇒ m = 0.0575 kg = 0.058 kg

(h)

τ = Fd = 410 × 0.55 = 225.5 Nm =230 Nm

Correct answer

(i)

The player must hold the racquet in such a way that an equal and opposite torque on the racquet is created.

Some idea of balanced torques

Correct answer

Idea of player creating equal and opposite torque

90258 • Demonstrate understanding of physics in an integrated context • Assessment Schedule 2007

3

Question

Evidence

Two (a)

The force on the electrons all around the loop will be at right angles to the magnetic field. This means that along QR the electrons will be pushed from one side of the wire to the other side, not along the wire. Therefore there will be no separation of charge along the wire and so no voltage across the length of the wire.

(b)

The magnetic field direction is the same for both PQ and RS. The direction of motion of PQ is opposite to RS. This means that whichever direction the electrons are pushed in PQ they will be pushed in the opposite direction in RS. This means that if Q has a positive potential R will be negative and this is the condition for voltages in series.

Idea of electrons being pushed in the same direction around the loop by both forces.

(c)

V = BvL = 0.75 × 0.23 × 2 × 0.058 = 0.02001 = 0.020 V

Correct answer for one side only

(d)

(f)

(g)

Correct answer consistent with length QR not being halved

mv 2 = r 9.11 ×10 −31 × 0.23 2 0.5 × 0.018

Fc =

−

5.35466× 10 30

= (e)

Achievement

V R

IR1 =

=

VR 1 R1

⇒I =

3.0 29 .2

Excellence

Idea of electrons not being pushed down the wire. (Do not accept no push on electrons.)

Idea of force on electrons across the wire meaning that there is no movement of electrons along the wire and hence no voltage.

Idea of electrons being pushed in the same direction around the loop by both forces is linked to the opposite direction of movement of the two sides of the loop. Correct answer

Correct answer

−

= 5.4× 10 30 N

Rparallel = Rtotal − R = 29.2 – 1.3 = 27.9 Ω

I=

Merit

= 0.10274

VR1 = 3.0 − Vr

= 0.10 A

Vr = IR

Correct answer

Correct answer

Correct concept for calculating VR1

Correct VR1

Correct answer

3.0 − (0.10274 × 1.3) 31

= 0.092466

= 0.092A

OR IR1 =

VR1 R1

VR1 = Vparallel = Itotal × Rparallel

= 0.10274 × 27.9 ⇒ IR1 = =

2.8664 31

= 2.8664

= 0.092465

0.092A

90258 • Demonstrate understanding of physics in an integrated context • Assessment Schedule 2007

4

Question

Evidence

1

Two (h)

R parallel ⇒

=

Achievement

1 1 + R 2 R1

1 1 1 = − R 2 27 .9 31

(j)

The wavelength of light is much shorter than the wavelength of radio waves. As amount waves can diffract depends on their wavelength, the radio waves are able to diffract around the upper part of the stand much more than the light waves and are able to reach the receiver.

Question

Evidence

Three (a)

1 1 1 = + f do di

⇒

⇒ di= 0.151515 (b)

npsinθ θ

Excellence

Correct answer

⇒ R2 = 279 = 280 Ω If R1 is reduced the total resistance in the circuit is reduced and so the current drawn from the battery is increased. This means that the voltage across R is also increased.

(i)

Merit

p

p

= nasinθ

1 sin 45 

Knowledge that wavelength is the key concept.

Achievement

Explanation is clear and complete.

Merit

Excellence

Correct answer

1 1 1 = − di 0.15 15

Correct answer

a a

= 90°

= 1.4142

= 1.4

(c)

Magnified image always produced.

Correct answer

(d)

Because the image is virtual the image distance is negative.

Recognition that the image distance is negative.

1 1 1 = − f do d i

Idea of changing the resistance changing the current from the battery and hence the voltage across R. Link made between longer wavelength and more diffraction.

= 0.15 m

= 45°, na = 1, θ

np =

Idea of reducing R1 reducing the total resistance in the circuit.

m=

di do

Correct answer consistent with a positive di being used.

Correct answer

1 1 1 = − f do 8do 1 7 = ⇒ 0.022 8d o ⇒

⇒ do = 0.01925

= 19 mm

90258 • Demonstrate understanding of physics in an integrated context • Assessment Schedule 2007

5 Grade Distribution: Explain Type (10 questions) Calculate (14 questions)

A u&sf 1b 1d 1e 2b 2j 3c

M 1b 1d 1i 2a 2b 2i 2j

E 1d 2a 2i

1f 1h 2e 2f

1a 1c 2c 2d 2h 3a 3b

1g 2g 3d

Sufficiency Statement Achievement: 8 correct answers Merit 12 correct answers of which 6 must be at the merit level or above Excellence: 12 correct answers of which 8 must be at the merit level or above and of these 8, 3 must be at the excellence level Or an alternative but equivalent format: Total Opportunities at each level:

Total : 11

Total : 14

Total : 6

Sufficiency

8

6A + 6M

4A+5M+3E

90258 • Demonstrate understanding of physics in an integrated context • Assessment Schedule 2007